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math110s-hw8sol

math110s-hw8sol - MATH 110 LINEAR ALGEBRA SPRING 2007/08...

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MATH 110: LINEAR ALGEBRA SPRING 2007/08 PROBLEM SET 8 SOLUTIONS 1. Let V be a vector space over F and S , T ∈ End( V ). (a) Show that I - S ◦ T is injective iff I - T ◦ S is injective. Solution. The trick is to observe that S ◦ ( I - T ◦ S ) = ( I - S ◦ T ) ◦ S . Suppose I - S ◦ T is injective. Let v ker( I - T ◦ S ). Then ( I - T ◦ S )( v ) = 0 V . Hence ( I - S ◦ T )( S ( v )) = S (( I - T ◦ S )( v )) = S ( 0 V ) = 0 V and since I - S ◦ T is injective, we deduce that S ( v ) = 0 V . So T ◦ S ( v ) = T ( 0 V ) = 0 V ; and so v = v - 0 V = v - T ◦ S ( v ) = ( I - T ◦ S )( v ) = 0 V where the last equality follows from our choice of v ker( I -T ◦S ). Hence ker( I -T ◦S ) = { 0 V } and we deduce that I - T ◦ S is injective by Theorem 4.12 . (b) T is called nilpotent if T n = O for some n N . Show that if T is nilpotent, then I - T is bijective. What is ( I - T ) - 1 ? Solution. Let S = I + T + T 2 + · · · + T n - 1 (by the geometric series heuristic in lecture). Then ( I - T ) ◦ S = ( I + T + · · · + T n - 1 ) - ( T + T 2 + · · · + T n ) = I - T n = I - O = I and likewise for S ◦ ( I - T ). Hence ( I - T ) - 1 = S and I - T is bijective. 2. Let V be a vector space over R and T ∈ End( V ) be an involution, ie. T 2 = I . Define V + := { v V | T ( v ) = v } and V - := { v V | T ( v ) = - v } . (a) Show that V + and V - are subspaces of V . Solution. Let α 1 , α 2 R and v 1 , v 2 V + . Then T ( v 1 ) = v 1 , T ( v 2 ) = v 2 , and T ( α 1 v 1 + α 2 v 2 ) = α 1 T ( v 1 ) + α 2 T ( v 2 ) = α 1 v 1 + α 2 v 2 . Hence α 1 v 1 + α 2 v 2 V + . Likewise for V - . Date : May 7, 2008 (Version 1.0). 1

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(b) Show that V + V - = V. Solution. Note that v = 1 2 ( I + T )( v ) + 1 2 ( I - T )( v ) . (2.1) Since T 2 = I , T ( 1 2 ( I + T )( v ) ) = T ( 1 2 v + 1 2 T ( v ) ) = 1 2 T ( v ) + 1 2 T 2 ( v ) = 1 2 T ( v ) + 1 2 I ( v ) = 1 2 ( I + T )( v ) and likewise, T ( 1 2 ( I - T )( v ) ) = T ( 1 2 v - 1 2 T ( v ) ) = 1 2 T ( v ) - 1 2 T 2 ( v ) = 1 2 T ( v ) - 1 2 I ( v ) = - 1 2 ( I - T )( v ) . Hence 1 2 ( I + T )( v ) V + , 1 2 ( I - T )( v ) V - , and (2.1) implies that V = V + + V - . If v V + V - , then v = T ( v ) = - v implies that v = 0 V . Hence V + V - = { 0 V } and so V = V + V - . 3. Let A R m × n and b R m . (a) Show that nullsp( A > A ) = nullsp( A ) . Solution. Let x nullsp( A ). Then A x = 0 . Therefore, A > A x = A > 0 = 0 . Therefore x nullsp( A > A ). So nullsp( A ) nullsp( A > A ). Conversely, let x nullsp( A > A ). Then A > A x = 0 . Multiplying on the left by x > , we get x > A > A x = x > 0 = 0 .
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